Chọn C.

Ta có tan α – cotα = 1 

Do  suy ra tanα < 0 nên 

Thay

 và

vào P  ta được 

Chọn D.

Ta có  suy ra : 

Thay  vào P ta được 

Chọn A.

Ta có 

Thay  vào P ta được P = 4.

D

D

a: \(\sin a+cosa=\sqrt2\)

=>\(\sqrt2\cdot\sin\left(a+\frac{\pi}{4}\right)=\sqrt2\)

=>\(\sin\left(a+\frac{\pi}{4}\right)=1\)

=>\(a+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
=>\(a=\frac{\pi}{4}+k2\pi\)

\(cosa=cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\sin a=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\)

\(\tan a=\tan\left(\frac{\pi}{4}\right)=1\)

\(\cot a=\cot\left(\frac{\pi}{4}\right)=1\)

b: \(F=\sin^5a+cos^5a\)

\(=\sin^5\left(\frac{\pi}{4}\right)+cos^5\left(\frac{\pi}{4}\right)=\left(\frac{\sqrt2}{2}\right)^5+\left(\frac{\sqrt2}{2}\right)^5\)

\(=\frac{4\sqrt2}{32}+\frac{4\sqrt2}{32}=\frac{8\sqrt2}{32}=\frac{\sqrt2}{4}\)

\(tan\left(\dfrac{3\pi}{2}-\alpha\right)+cot\left(3\pi-\alpha\right)-cos\left(\dfrac{\pi}{2}-\alpha\right)+2.sin\left(\pi+\alpha\right)\)

\(=tan\left(\pi+\dfrac{\pi}{2}-\alpha\right)+cot\left(-\alpha\right)-sin\alpha+2\left(sin\pi.cos\alpha+cos\pi.sin\alpha\right)\)

\(=tan\left(\dfrac{\pi}{2}-\alpha\right)-cot\alpha-sin\alpha+2.-sin\alpha\)

\(=cot\alpha-cot\alpha-3sin\alpha\)

\(=-3sin\alpha\)

Chọn A.

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

3/4pi<a<pi

=>sin a>0; cosa<0

sin2a=-4/5

=>2*sina*cosa=-4/5

=>sina*cosa=-2/5

(sina-cosa)^2=sin^2a+cos^2a-2*sina*cosa=1+4/5=9/5

=>sin a-cosa=3/căn 5